Similar Polygons. These rectangles are not similar. In the investigation, you will explore what makes polygons similar.

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1 CONDENSED LESSON 11.1 Similar Polygons In this lesson, you Learn what it means for two figures to be similar Use the definition of similarity to find missing measures in similar polygons Explore dilations of figures on a coordinate plane You know that figures that have the same size and shape are congruent figures. Figures that have the same shape but not necessarily the same size are similar figures. You can think of similar figures as enlargements or reductions of one another with no distortion. The pentagons below are similar. The rectangles are not similar. You could not enlarge or reduce one rectangle to fit exactly over the other. These pentagons are similar. These rectangles are not similar. In the investigation, you will explore what makes polygons similar. Investigation 1: What Makes Polygons Similar? Hexagons ABCDEF and PQRSTU in your book are similar. Use patty paper to compare the corresponding angles. You should find that the corresponding angles are congruent. Now measure the corresponding sides, and find the ratio of each side length of the smaller hexagon to the corresponding side length of the larger hexagon. You should find that each ratio is equal to 5. 9 So, the corresponding sides are proportional. The relationships you discovered in the investigation illustrate the mathematical definition of similar polygons: Two polygons are similar if and only if the corresponding angles are congruent and the corresponding sides are proportional. The statement CORN PEAS says that quadrilateral CORN is similar to quadrilateral PEAS. The order of the letters tells you which sides and angles correspond. Look at quadrilaterals SQUE and RCTL on page 565 of your book. These figures have congruent corresponding angles, but they are not similar because the corresponding sides are not proportional. Now, look at quadrilaterals SQUE and RHOM. These figures have corresponding sides that are proportional, but they are not similar because the corresponding angles are not congruent. (continued) Discovering Geometry Condensed Lessons CHAPTER 11 14

2 Lesson 11.1 Similar Polygons (continued) These examples illustrate that for two figures to be similar both conditions proportional sides and congruent angles must hold. Here is another example. EXAMPLE Determine whether parallelogram MNOP is similar to parallelogram WXYZ. W M P 8 cm 6 cm N 8 cm O X 12 cm Y Z Solution mn mx. Using angle properties of parallelograms, mm mw 120, mp mz 60, and mo my 120. So, the corresponding angles are congruent. However, because M N WX and N O 8 XY 1 2 2, the corresponding sides are not proportional. Therefore, the parallelograms are not similar. If you know that two polygons are similar, you can use the definition of similar polygons to find missing measures. The example in your book shows you how. Read this example carefully and make sure you understand it. In an earlier chapter, you saw that rigid transformations translations, rotations, and reflections preserve the size and shape of a figure, resulting in an image that is congruent to the original. In the next investigation, you will look at a nonrigid transformation called a dilation. Investigation 2: Dilations on a Coordinate Plane To dilate a figure in a coordinate plane, you multiply all its coordinates by the same number, called a scale factor. Pentagon ABCDE in your book has vertices with coordinates A(4, 4), B(2, 6), C(4, 4), D(6, 2), E(0, 6). If you multiply each coordinate by 1, 2 you get A(2, 2), B(1, ), C(2, 2), D(, 1), E(0, ). The figure at right shows the original pentagon and the image. B B 7 y C C If you compare the sides and angles, you will find that the 7 corresponding angles are congruent and that each side of the A image pentagon is half the length of the original side. So, the A pentagons are similar. Choose one of the other scale factors listed in your book 4,2, 7 or and multiply each coordinate of ABCDE by the factor you choose. Graph the image pentagon on the same axes as the original. Compare the corresponding angle measures and side lengths. Are the original and the image similar? State your findings as a conjecture. E E D 7 D x Dilation Similarity Conjecture If one polygon is the image of another polygon under a dilation, then. C CHAPTER 11 Discovering Geometry Condensed Lessons

3 CONDENSED LESSON 11.2 Similar Triangles In this lesson, you Learn shortcuts for determining whether two triangles are similar In Lesson 11.1, you saw that, to determine whether two quadrilaterals are congruent, you must check both that their sides are proportional and that their angles are congruent. 6 in. 9 in. Angles are congruent, but quadrilaterals are not similar. 10 in. Sides are proportional, but quadrilaterals are not similar. 15 in. However, triangles are different. In Chapter 4, you discovered that you don t have to check every pair of sides and angles to determine whether two triangles are congruent. You found that SSS, SAS, ASA, and SAA are congruence shortcuts. In this lesson, you will determine whether there are similarity shortcuts. Page 572 of your book shows two triangles in which one pair of corresponding angles is congruent. The triangles are clearly not similar. So, knowing only that one pair of angles is congruent is not enough to conclude that two triangles are similar. What if two pairs of angles are congruent? Investigation 1: Is AA a Similarity Shortcut? In the triangles at right, A D and B E. What must be true about C and F? Why? Measure the sides and compare the ratios of corresponding side lengths. AC C D? Is D AB E F B EF A B D E Now, draw your own triangle ABC. Use a compass and straightedge to construct a second triangle, DEF, with A D and B E. Are your triangles similar? Explain. Your findings should support this conjecture. C F AA Similarity Conjecture If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. C-9 The illustration on page 57 of your book shows that you cannot conclude that two triangles are similar given that two pairs of corresponding sides are proportional. What if three pairs of corresponding sides are proportional? (continued) Discovering Geometry Condensed Lessons CHAPTER

4 Lesson 11.2 Similar Triangles (continued) Investigation 2: Is SSS a Similarity Shortcut? Use the segments on the left to construct ABC. Each segment on the right is three times the length of the corresponding segment on the left. Use the segments on the right to construct DEF. A B D E B C E F A C D F The sides of DEF are proportional to the sides of ABC. Measure the angles and see how they compare. Are the triangles similar? Construct another pair of triangles so that the side lengths of one triangle are a multiple of the side lengths of the other. Compare the corresponding angles of your triangles. Your work in this investigation should support this conjecture. SSS Similarity Conjecture If the three sides of one triangle are proportional to the three sides of another triangle, then the triangles are similar. C-94 If AA is a similarity shortcut, then so are ASA, SAA, and AAA, because each of those shortcuts contains two angles. That leaves SAS and SSA as possible shortcuts. In the next investigation, you will look at SAS. Investigation : Is SAS a Similarity Shortcut? Try to construct two different triangles that are not similar, but that have two pairs of sides proportional and the pair of included angles equal in measure. Can you do it? Your findings should support this conjecture. SAS Similarity Conjecture If two sides of one triangle are proportional to two sides of another triangle and the included angles are congruent, then the triangles are similar. C-95 In the triangles below, two pairs of corresponding sides are proportional and one pair of non-included angles is congruent. However, the triangles are clearly not similar. This shows that SSA is not a similarity shortcut. B A 24 cm 16 cm 10 cm 0 0 C E F 15 cm D AB 16 2_ DE 24 AC 10 2_ DF 15 AB AC D E D F and B E, but the triangles are not similar. 146 CHAPTER 11 Discovering Geometry Condensed Lessons

5 CONDENSED LESSON 11. Indirect Measurement with Similar Triangles In this lesson, you Learn how to use similar triangles to find heights of tall objects indirectly Suppose you want to find the height of a tall object such as a flagpole. It would be difficult to measure the flagpole directly you would need a very tall ladder and a very long tape measure! In this lesson, you will learn how you can use similar triangles to find the heights of tall objects indirectly. Investigation: Mirror, Mirror You will need another person to help you with this investigation. You will also need a small mirror. Mark crosshairs on your mirror with masking tape or a soluble pen. Label the intersection point X. Choose a tall object, such as a flagpole, a tall tree, a basketball hoop, or a tall building. Set the mirror, faceup, on the ground several yards from the object you wish to measure. Step back from the mirror, staying in line with the mirror and the object, until you see the reflection of the top of the object at point X on the mirror. Have another person measure and record the distances from you to X and from X to the base of the object. Also, have the person measure and record your height at eye level. Sketch a diagram of your setup, like this one. Label the top of the object T, the base of the object B, the point where you stood P, and your eye level E. Label PX, BX, and EP with the measurements your helper found. E Think of TX as a light ray that bounced back to your eye along XE. Because the incoming angle must be congruent to the outgoing P angle, EXP TXB. Also, because EP and TB are perpendicular X to the ground, P B. By the AA Similarity Conjecture, EPX TBX. Because the triangles are similar, you can set up a proportion to find TB, the height of the tall object. EP TB P X B X X T B Find the height of your object. Then, write a paragraph summarizing your work. Discuss possible causes for error. (continued) Discovering Geometry Condensed Lessons CHAPTER

6 Lesson 11. Indirect Measurement with Similar Triangles (continued) The example in your book illustrates a method of indirect measurement that involves shadows. Read the example carefully. Make sure you can explain why the two triangles are similar. Here is another example. EXAMPLE A brick building casts a shadow 28 feet long. At the same time, a -foot-tall child casts a shadow 6 inches long. How tall is the building? Solution The drawing below shows the similar triangles formed. Find x by setting up and solving a proportion. x ft 0.5 ft 28 ft Height of child Length of child s shadow 0.5 x 28 x x 168 x The building is 168 feet tall. Height of building Length of building s shadow 148 CHAPTER 11 Discovering Geometry Condensed Lessons

7 CONDENSED LESSON 11.4 Corresponding Parts of Similar Triangles In this lesson, you Investigate the relationship between corresponding altitudes, corresponding medians, and corresponding angle bisectors of similar triangles Prove that corresponding medians of similar triangles are proportional to corresponding sides Discover a proportional relationship involving angle bisectors If two triangles are similar, then their side lengths are proportional. In the next investigation, you will see if there is a relationship between corresponding altitudes, corresponding medians, or corresponding angle bisectors. Investigation 1: Corresponding Parts PQR DEF. The scale factor from PQR to DEF is 4. P M D N H Q A R J E B F PH and DJ are corresponding altitudes. How do the lengths of these altitudes compare? How does the comparison relate to the scale factor? PA and DB are corresponding medians. How do the lengths of these medians compare? QM and EN are corresponding angle bisectors. How do the lengths of these angle bisectors compare? Now, draw your own triangle and then construct a similar triangle of a different size. Tell what scale factor you used. Follow Steps 2 4 in your book to construct and compare corresponding altitudes, medians, and angle bisectors. Summarize your findings in this investigation by completing the conjecture below. Proportional Parts Conjecture If two triangles are similar, then the corresponding,, and are to the corresponding sides. C-96 The example in your book proves that the corresponding medians of similar triangles are proportional to the corresponding sides. Read the example carefully, following along with each step. (continued) Discovering Geometry Condensed Lessons CHAPTER

8 Lesson 11.4 Corresponding Parts of Similar Triangles (continued) If a triangle is isosceles, the bisector of the vertex angle divides the opposite sides into equal parts. (That is, the angle bisector is also a median.) However, as the triangle on the right below shows, this is not always true. A D B X AX is an angle bisector. Point X is the midpoint of BC. C E Y M DY is an angle bisector. Point M is the midpoint of EF. F The angle bisector does, however, divide the opposite side in a particular way. Investigation 2: Opposite Side Ratios Follow Steps 1 in your book. Measure and compare CD and BD. Calculate and compare the ratios C A BA and C D BD. You should find that both ratios are equal to 1 2. Repeat Steps 1 5 in your book with AC 6 units and AB 18 units. This time you should find that C A BA and C D BD are both equal to 1. You can state your findings as a conjecture. Angle Bisector/Opposite Side Conjecture A bisector of an angle in a triangle divides the opposite side into two segments whose lengths are in the same ratio as the lengths of the two sides forming the angle. C CHAPTER 11 Discovering Geometry Condensed Lessons

9 CONDENSED LESSON 11.5 Proportions with Area and Volume In this lesson, you Discover the relationship between the areas of similar figures Discover the relationship between the volumes of similar solids How does multiplying the dimensions of a two-dimensional figure by a scale factor affect its area? How does multiplying the dimensions of a three-dimensional solid by a scale factor affect its volume? In this lesson, you will explore these questions. Investigation 1: Area Ratios The rectangle on the right was created by multiplying the side lengths of the rectangle on the left by. 1 9 The area of the small rectangle is square units. The area of the large rectangle is 27 square units. The ratio of side lengths of the larger rectangle to side lengths of the smaller rectangle is 1, and the ratio of areas is 9. 1 Notice that nine copies of the small rectangle fit inside the large rectangle. 1 9 Now, draw your own rectangle on graph paper. Then, create a larger or smaller rectangle by multiplying the sides by a scale factor other than. What is the ratio of side lengths of the larger rectangle to side lengths of the smaller rectangle? What is the ratio of the areas? How do the two ratios compare? Draw a triangle on graph paper (your work will be easiest if you draw a right triangle). Draw a similar triangle by multiplying the side lengths by a scale factor. Find the ratio of side lengths and the ratio of areas. How do the ratios compare? Do you think your findings would be the same for any pair of polygons? Would your findings be the same for a circle or other curved figure? For example, consider a circle with radius 5 cm and a circle with radius 20 cm. How does the ratio of radii compare to the ratio of areas? (continued) Discovering Geometry Condensed Lessons CHAPTER

10 Lesson 11.5 Proportions with Area and Volume (continued) The following conjecture summarizes the relationship between the areas of similar figures. Proportional Areas Conjecture If corresponding sides of two similar polygons or the radii of two circles compare in the ratio m, n then their areas compare in the ratio m 2 n. 2 C-98 Similar solids are solids that have the same shape but not necessarily the same size. All cubes are similar, but not all prisms are similar. All spheres are similar, but not all cylinders are similar. Two polyhedrons are similar if all their corresponding faces are similar and the lengths of their corresponding edges are proportional. Two right cylinders or right cones are similar if their radii and heights are proportional. Examples A and B in your book involve determining whether two given solids are similar. Try to answer the problems yourself before reading the solutions. Investigation 2: Volume Ratios In this investigation, you ll explore how the ratio of the edge lengths of similar solids compares to the ratio of the volumes. You will need interlocking cubes. Follow Steps 1 and 2 in your book. In Step 2, make sure you multiply all three dimensions length, width, and height by. What is the ratio of the side lengths (larger to smaller) for the two snakes? What is the ratio of the volumes? How do the ratios compare? Follow Step 4 in your book. How does multiplying each dimension by 2 affect the volume? How would the volume change if you multiplied each dimension by 4? How would the volume change if you multiplied each dimension by 5? Your findings can be stated as a conjecture. Proportional Volumes Conjecture If corresponding edges (or radii, or heights) of two similar solids compare in the ratio m, n then their volumes compare in the ratio m n. C CHAPTER 11 Discovering Geometry Condensed Lessons

11 CONDENSED LESSON 11.6 Proportional Segments Between Parallel Lines In this lesson, you Explore the relationships in the lengths of segments formed when one or more lines parallel to one side of a triangle intersect the other two sides Learn how you can use the relationship you discover to divide a given segment into any number of equal parts The top of page 60 in your book shows LUV and line MT, with MT LU. It appears that LUV MTV. The paragraph proof given uses the AA Similarity Conjecture to prove this is true. Example A in your book uses this idea to solve a problem. Read the example, and follow along with each step. LE Look back at the figure from Example A. Notice that E 45 and NO O M M 60 4, so there are more relationships in the figure than the ones found by using similar triangles. You will explore these relationships in the next investigation. Investigation 1: Parallels and Proportionality Step 1 of the investigation gives three triangles, each with a line parallel to one side. For each triangle, find x and then find the values of the specified ratios. Here is the answer to part a. a. Use the fact that CDE BDA to write and solve a proportion. D E D A B Corresponding parts of similar triangles are proportional x Substitute the lengths from the figure x Simplify the left side. 12 x 6 Multiply both sides by (12 x) and simplify. x 24 Subtract 12 from both sides. So, D E 8 AE 1 6 C and D BC 1 2. In each part of Step 1, you should find that the ratios of the lengths of the segments cut by the parallel lines are equal. In other words: If a line parallel to one side of a triangle passes through the other two sides, then it divides the other two sides proportionally. Do you think the converse of this statement is also true? That is, if a line divides two sides of a triangle proportionally, is it parallel to the third side? Follow Steps 7 in your book. You should find that PAC PBD, so AC and BD are parallel. PA PC Repeat Steps 7, but this time mark off your own lengths, such that A B C D. Again, you should find that AC is parallel to BD. You can use your findings from this investigation to state a conjecture. (continued) Discovering Geometry Condensed Lessons CHAPTER 11 15

12 Lesson 11.6 Proportional Segments Between Parallel Lines (continued) Parallel/Proportionality Conjecture If a line parallel to one side of a triangle passes through the other two sides, then it divides the other two sides proportionally. Conversely, if a line cuts two sides of a triangle proportionally, then it is parallel to the third side. C-100 Example B proves the first part of the Parallel/Proportionality Conjecture. Try to write a proof yourself before reading the one in the book. Use the fact that AXY ABC to set up a proportion. Then, write a series of algebraic steps until you get a c b. d Investigation 2: Extended Parallel/Proportionality In the triangles in Step 1, more than one segment is drawn parallel to one side of a triangle. Find the missing lengths. Here is the solution to part a. To find x and y, apply the Parallel/Proportionality Conjecture to the appropriate triangles and lines. a. To find x, use AEL and FT. E F ET FL T A x x 70 To find y, use REG and LA. EL EA LG A R y y 56 F Using the values of x and y, you can see that L G The results of Step 1 lead to the following conjecture. L TA A R 5 4. Extended Parallel/Proportionality Conjecture If two or more lines pass through two sides of a triangle parallel to the third side, then they divide the two sides proportionally. C-101 You can use the Extended Parallel/Proportionality Conjecture to divide a segment into any number of equal parts. Example C in your book shows you how to divide a segment, AB, into three equal parts. Read the example carefully. To make sure you understand the process, divide XY below into four equal parts using a compass and straightedge. X Y 154 CHAPTER 11 Discovering Geometry Condensed Lessons

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